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Bunuel
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I'm confused why the answer is not (E).

From Statement (1):
q(r+s) = r +s
Therefore q = 1 or 0

From Statement (2):
We know (r+s) ≠ 0

Statements (1) + (2) - couldn't variable q still be 1 or 0?

Edit: Annnnnd, I just realized that q cannot be 0 because it were, statement (1) would not hold true. Too quick to post.
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What is the value of integer q?

(1) qr + qs = r + s
(2) r ≠ -s

Required: q = ?

Statement 1: qr + qs = r + s
Or q (r + s) = (r+s)
(q-1)(r+s) = 0 (i)
Either q = 1 or r = -s
INSUFFICIENT

Statement 2: r ≠ -s (ii)
Nothing about q.
INSUFFICIENT

Combining Statement 1 and Statement 2:
From (i), we have (q-1)(r+s) = 0 (i)
And from (ii) r ≠ -s

Hence q = 1
SUFFICIENT

Option C



How did you get the colored step?
How did it become (q-1)
And can someone explain why can't we do this

qr + qs = r + s

q(r+s)= r +s
q=\(\frac{(r+s)}{(r+s)}\)

divide (r+s) on both sides of the equation. The answer will be q =1 Why is this step wrong here?
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Bunuel
What is the value of integer q?

(1) qr + qs = r + s
(2) r ≠ -s

Required: q = ?

Statement 1: qr + qs = r + s
Or q (r + s) = (r+s)
(q-1)(r+s) = 0 (i)
Either q = 1 or r = -s
INSUFFICIENT

Statement 2: r ≠ -s (ii)
Nothing about q.
INSUFFICIENT

Combining Statement 1 and Statement 2:
From (i), we have (q-1)(r+s) = 0 (i)
And from (ii) r ≠ -s

Hence q = 1
SUFFICIENT

Option C



How did you get the colored step?
How did it become (q-1)
And can someone explain why can't we do this

qr + qs = r + s

q(r+s)= r +s
q=\(\frac{(r+s)}{(r+s)}\)

divide (r+s) on both sides of the equation
. The answer will be q =1 Why is this step wrong here?

Hi mtk10

The highlighted portion is not correct. r & s are variables whose value you don't know.

Suppose r=2 & s=-2, then

qr + qs = r + s =>2q-2q=2-2=0. Now q can be any integer, not necessarily 1

NEVER CANCEL OUT VARIABLES

so the equation will be q(r+s)=(r+s) => q(r+s)-(r+s)=0

(r+s)(q-1)=0 so Either r+s=0 or q=-1. Hence the statement is not sufficient
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Bunuel
What is the value of integer q?

(1) qr + qs = r + s
(2) r ≠ -s

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (q, r and s) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
qr+qs = r+s
⇔ q(r+s) = r+s
⇔ q(r+s) - (r+s) = 0
⇔ (q-1)(r+s) = 0
⇔ q-1=0 or r+s = 0
⇔ q-1=0 since r≠-s
⇔ q = 1.

Both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
q = 1, r = 1, s = 1
q = 2, r = 1, s = -1

Since we don't have a unique solution, condition 1) is not sufficient.

Condition 2)
Since we don't have information about q, condition 2) is not sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Bunuel
What is the value of integer q?

(1) qr + qs = r + s
(2) r ≠ -s

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (q, r and s) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
qr+qs = r+s
⇔ q(r+s) = r+s
⇔ q(r+s) - (r+s) = 0
⇔ (q-1)(r+s) = 0
⇔ q-1=0 or r+s = 0
⇔ q-1=0 since r≠-s
⇔ q = 1.

Both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
q = 1, r = 1, s = 1
q = 2, r = 1, s = -1

Since we don't have a unique solution, condition 1) is not sufficient.

Condition 2)
Since we don't have information about q, condition 2) is not sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

Quant Newbie here and I dont understand how to get to the highlighted step, can someone explain? :?
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Bunuel
What is the value of integer q?

(1) qr + qs = r + s
(2) r ≠ -s

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (q, r and s) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
qr+qs = r+s
⇔ q(r+s) = r+s
⇔ q(r+s) - (r+s) = 0
⇔ (q-1)(r+s) = 0
⇔ q-1=0 or r+s = 0
⇔ q-1=0 since r≠-s
⇔ q = 1.

Both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
q = 1, r = 1, s = 1
q = 2, r = 1, s = -1

Since we don't have a unique solution, condition 1) is not sufficient.

Condition 2)
Since we don't have information about q, condition 2) is not sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

Quant Newbie here and I dont understand how to get to the highlighted step, can someone explain? :?

It's done by factoring out r + s from q(r + s) - (r + s) = 0:

(r + s)(q - 1) = 0.

Does this makes sense?
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Bunuel
What is the value of integer q?

(1) qr + qs = r + s
(2) r ≠ -s

Hey everyone,
On solving s1 we get (q-1)(r+s)
now, two solutions are possible whether => q=1 or r=-s

But as in the question, only the value q is asked
why do we need s2 to support s1.
S2 just tells that r=-s is not a solution.

how does this affect the value of q? As the value of will remain same

Bunuel chetan2u can you help guys?
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Ahmed9955
Bunuel
What is the value of integer q?

(1) qr + qs = r + s
(2) r ≠ -s

Hey everyone,
On solving s1 we get (q-1)(r+s)
now, two solutions are possible whether => q=1 or r=-s

But as in the question, only the value q is asked
why do we need s2 to support s1.
S2 just tells that r=-s is not a solution.

how does this affect the value of q? As the value of will remain same

Bunuel chetan2u can you help guys?

Yes, from (r + s)(q - 1) = 0 it follows that r = -s OR q = 1. But notice that if q = -s, then q is not necessarily 1, it can be any other number too.
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